= Solution
The <tower property of conditional expectation> gives
$$
\mathbb E\!\left[\frac{YT}{\pi(X)}\middle|X\right]
=\frac{\mathbb E[YT\mid X]}{\pi(X)}
=\frac{\pi(X)\mu(X)}{\pi(X)}=\mu(X),
$$
and another expectation gives $\theta$.
Use the <augmented inverse-probability-weighted estimator>
$$
\widehat\theta_n=\frac1n\sum_{i=1}^n
\left[\widehat\mu_n(X_i)+\frac{T_i}{\widehat\pi_n(X_i)}
\{Y_i-\widehat\mu_n(X_i)\}\right].
$$
Condition on the independently trained nuisance estimators. Subtracting the oracle influence variable
$$
\phi(X,Y,T)=\mu(X)+\frac{T}{\pi(X)}(Y-\mu(X))
$$
produces a conditional empirical fluctuation with variance $o(1)$ after multiplication by $\sqrt n$, using $\mathcal E_\mu,\mathcal E_\pi\to0$, overlap, and the bounded conditional variance. Its conditional bias is
$$
\mathbb E\left[(\widehat\mu_n(X)-\mu(X))
\left(1-\frac{\pi(X)}{\widehat\pi_n(X)}\right)\right],
$$
whose absolute value is at most $\sqrt{\mathcal E_\mu\mathcal E_\pi}=o(n^{-1/2})$ by <Cauchy-Schwarz inequality>. Thus
$$
\sqrt n(\widehat\theta_n-\theta)
=\frac1{\sqrt n}\sum_{i=1}^n(\phi_i-\theta)+o_P(1).
$$
The <central limit theorem> and <Slutsky theorem> give the claimed $N(0,v)$ limit. Without auxiliary data, use <cross-fitting>: split the sample into folds, train both nuisance estimators away from each observation's fold, and average the same score over held-out observations.
Solved by gpt-5.6-sol high.
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